CODEWITHSUNDEEP
cstringpointers
Dunc
Dunc

Reputation: 8058

Array of char pointers in C - Distinguishing empty elements

I have an array of char pointers in C with a size of 10.

I am trying to loop through and print each string out. Not all of the elements are populated and I believe this is causing me an error. How do I check if an element is populated.

This is what I have tried. When I run my code I find that the first element is populated and the value is printed but then I get an error.

char *errors[10];
ret = doMessages(&h, errors);
for (i = 0; i < 10; i++)
{
    if(errors[i] != NULL)
    {
        printf("%s", errors[i]);
    }
}

Upvotes: 1

Views: 2057

Answers (3)

Jon
Jon

Reputation: 437554

You do not explicitly initialize the values in your array to NULL, so they contain whatever happened to previously exist in the memory location where your array is stored.

To initialize, use something like

char *errors[10] = { 0 };

You could also do the same with memset:

char *errors[10];
memset(errors, 0, sizeof(errors));

Upvotes: 0

asaelr
asaelr

Reputation: 5456

Well, probably only the first element of errors is been populated. The rest are garbage values (may be NULL, and may be anything), so your check (!=NULL) doesn't really check that an element has populated.

I suggest you to initialize your array before doMessages, (something like char *errors[10]={0};, and then the check errors[i]!=NULL will make sense.

Upvotes: 0

hmjd
hmjd

Reputation: 122001

errors is not initialised so the elements can be any value, specifically may not be NULL. Change to:

char *errors[10] = { 0 }; /* Initialise all elements to NULL. */

Upvotes: 4

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